[Fpga-synth] Simple Filter Question

[Scott Gravenhorst]

Magnus Danielson <magnus at rubidium.dyndns.org> wrote:
>From: Scott Gravenhorst <music.maker at gte.net>
>Subject: [Fpga-synth] Simple Filter Question
>Date: Mon, 18 Jun 2007 13:35:14 -0700
>Message-ID: <200706182035.l5IKZEcr003383 at linux7.lan>
>
>Hi Scott,
>
>> The terrormouse PDF specifies that the filter used at the end of the delay
>> line is a "simple first order IIR".  It then gives this to describe it:
>> 
>>                        -1
>>        0.8995 + 0.1087z
>> H(z) = ------------------
>>                      -1
>>           1 + 0.0136z
>> 
>> >From what I understand, this makes:
>> 
>> a0 = 0.8995      a1 = 0.1087
>>                  b1 = -0.0136
>> 
>> There was no explanation for the values chosen except that they "sound like
>> an acoustic guitar string".  
>> 
>> What I don't understand is that this matches neither a first order lowpass
>> nor highpass.  Lowpass should have only a0 and b1 terms and highpass has a1
>> = - a0 (at least).
>> 
>> I am assuming this is supposed to be a lowpass filter of some kind.
>> 
>> When I did my testing, I used the first order lowpass as described at
>> www.dspguide.com, in the recursive filters PDF.  It's not the same filter,
>> but it works.
>> 
>> A resource that explains this would be most appreciated ... ?
>
>Will the list-admin do?
>
>Hmm...
>
>       0.8995 z + 0.1087
>H(z) = -----------------
>         z + 0.0136
>
>z_z = -0.1087/0.8995 = -0.1208
>z_p = -0.0136
>
>i.e.
>
>              z + 0.1208
>H(z) = 0.8995 ----------
>              z + 0.0136
>
>Now, what is this creature? Plot the pole and zero in the z-plane
>
>            
>         ^ jw
>         |
>         |         _
>         |        o
>---o---x-+-------->
>         |
>         |
>         |
>
>This is a clear case of a low-pass filter.
>
>For low frequency (i.e. z = 1) the pole is nearer than the zero, so the pole
>dominates and the gain for this case is 0.9947 (insert z = 1 in H(z)).
>As you move along the unit circle you come closer to the zero and at the
>Nyquist frequency (i.e. z = -1) the zero dominates and the gain is 0.8017
>(insert z = -1 in H(z)). This is similar to a shelving filter.
>Midways we have a gain of 0.9125 (insert z = j in H(z) and take absolute
>value of complex value) at half the Nyquist frequency.
>
>You can view this as a shelving filter. which lowers the high frequency with
>about -1.9 dB.
>
>Did that make any sense?

Yes, thanks, and most of it did.  What I needed to do is exactly what you did, derive
poles and zeroes and look at the Z plane as you did.  Once I saw that I did agree that it
has to be a lowpass.

I need to learn how you turned the H(z) form into the z-plane, but it doesn't look too
bad, at least for such a simple filter.  

What is this process called so I can read about it?




-- ScottG

-------------------------------------------------------------

-- Scott Gravenhorst
-- GateMan I - Xilinx Spartan-3E Based MIDI Synthesizer
-- FatMan: home1.gte.net/res0658s/fatman/
-- NonFatMan: home1.gte.net/res0658s/electronics/
-- When the going gets tough, the tough use the command line.